Skip to content

Applying Fujimoto approximation in folding Origami Spirals

Materials: Strips of paper with sides about 5x30cm long (which can be obtained by dividing the width of a piece of A4 paper into 4 equal parts), board, and writing markers.

Objectives: Students understand the principle of this approximation method, the bases of numbers, other parts of number theory, and the general algorithm, and then they can apply the acquired knowledge in folding origami spirals.

Topics: approximation, number bases, congruence, error.


  1. Introduce the challenge of dividing a strip of paper into n equal parts without using a ruler and simplify it by focusing on the scenario where n is odd. (You can achieve this by folding the paper in half repeatedly until you are left with an odd number of divisions). Begin by addressing the case of n = 5 and encourage students to come up with different algorithms to solve the problem.
  • Demonstrate the Fujimoto approximation in its simplest form by considering fractions like ½ or ⅓.

The Fujimoto approximation leverages the simplicity of halving as a fundamental operation. By recognizing this advantage, we can utilize it to our benefit.

a) To achieve the desired 1/n value, we can approximate and shift a section to the left.

b) The right section of the paper strip will constitute (n-1)/n of its total length. We can fold it in half.

c) Starting with a section of paper, each new crease divides it into two parts. We then halve the part containing an even number of nths. Repeat this process until we reach a part that is 1 nth long. As a result, this final part will be closer to the desired 1nth length than our initial estimation.

Questions: Can we know beforehand which side to fold, or how many times to fold? 

  • Ask students to explain why the algorithm works. Specifically, the example of n = 5 can be used to illustrate how the deviation is halved with each fold until it approaches zero.
  • Consider a strip with a length of 1 unit, and let the initial piece be ⅕ + e units long. After n folds, the deviation decreases to e/2^n units, where n represents the number of folds. For significantly large values of n, this deviation becomes insignificant.

Number bases: To better understand decimals, let’s take an example: 124.75 can be expressed as 1 * 10^2 + 2 * 10^1 + 4 * 10^0 + 7 * 10^(-1) + 5 * 10^(-2). Now, let’s move on to representing the fraction 1/5 in binary. After 8 digits, we observe that the sequence starts repeating. This holds true for any rational number in any base.

Delay the proof, and ask students whether they want a sketch of this solution.

124.75 = 1.2^6 + 1.2^5 + 1.2^4 + 1.2^3+1. 2^2 + 0.2^1 + 0.2^0 + 1.2^-1 + 1.2^-2 = 1111100.11 (binary)

  1. Instruct students on the concept of number bases and demonstrate it by writing numbers in base 2. Allude to the problem of why all rational numbers have periodic decimal representations.

Algorithm: To determine the appropriate folding direction (left or right) in a binary fraction, follow these steps:

  • Identify the repeating sequence in the binary representation of the fraction.
    • Reverse the sequence.
    • Fold from the right whenever a 1 is encountered.
    • Fold from the left when a 0 is reached. By adhering to this procedure, you can accurately determine when to fold from the right and when to fold from the left in the given binary fraction.

Justification: We shall employ the following functions to define the properties of the folds we create:

Function: T_0(x) = x/2 for even values of n.x, T_1(x) = (x+1)/2 for odd values of n.x

Procedure: When folding, apply function T_0 if the crease position x consists of even parts. Alternatively, use function T_1 if x consists of odd parts.

To clarify, we can represent the process as follows: T0 adds a 0 to the sequence, and T1 adds a 1. To ensure the digits are inserted correctly, we must add them to the end of the repeating sequence first. In other words, when performing the procedure, we need to add any new digits after the previous repeating digits. For example, if the repeating sequence is 0.011, and we apply T0, the result would be 0.0110. If we apply T1 instead, the result would be 0.0111. This ensures that the digits further down the decimal part are inserted before, hence the reversed order.

To understand why rational numbers have repeating decimal parts, consider the binary representation of 1/5. The n-th digit after the decimal point in this representation is the remainder obtained when dividing the floor of 2^n/5 by 2. To prove that the decimal part repeats, we need to show that the remainder sequence generated from this calculation repeats.

  1. Present the general Fujimoto approximation:

Typically, people struggle to achieve precise hand division of a piece into exactly 30 parts (including myself). However, this task should be approached as a fun activity rather than a strict requirement. Even dividing the piece into 25 parts or more using slightly thicker paper can already be considered a successful outcome.


  1. Students apply their acquired knowledge to solve the problem where n = 30 = 2x3x5

Begin by dividing the strip into equal rectangles, then proceed to mark all of the lower left-upper right or lower right-upper left diagonals on these rectangles. Afterward, perform a valley fold along the marked diagonals and a mountain fold along the straight lines that were previously divided.

Note: Please ensure that the procedure is explained thoroughly so that everyone understands it. This approach has been successfully applied to 11-15-year-olds by some individuals. Depending on your goal, you can either follow steps 1 to 3, which provide a reliable algorithm for approximation and then proceed to create the spiral. Alternatively, you can spend additional time exploring the different number-theoretical properties before moving on.

  •  Students use the strips of divided paper to create origami spirals, instructors monitor & help when needed.

Leave a Reply

Your email address will not be published. Required fields are marked *